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frosja888 [35]
1 year ago
11

Consider the following skeletal C program: void fun1(void); /* prototype */ void fun2(void); /* prototype */ void fun3(void); /*

prototype */ void main() { int a, b, c; . . . } void fun1(void) { int b, c, d; . . . } void fun2(void) { int c, d, e; . . . } void fun3(void) { int d, e, f; . . . } 256 Chapter 5 Names, Bindings, and Scopes Given the following calling sequences and assuming that dynamic scoping is used, what variables are visible during execution of the last function called? Include with each visible variable the name of the function in which it was defined. a. main calls fun1; fun1 calls fun2; fun2 calls fun3. b. main calls fun1; fun1 calls fun3. c. main calls fun2; fun2 calls fun3; fun3 calls fun1. d. main calls fun3; fun3 calls fun1. e. main calls fun1; fun1 calls fun3; fun3 calls fun2. f. main calls fun3; fun3 calls fun2; fun2 calls fun1
Computers and Technology
1 answer:
natita [175]1 year ago
8 0

Answer:

Check the explanation

Explanation:

a) main calls fun1; fun1 calls fun2; fun2 calls fun3

fun3()                                        d, e, f

fun2()                                        c, d, e

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "c" of fun2() will be visible

   variable "b" of fun1() will be visible

   variable "a" of main() will be visible

b) main calls fun1; fun1 calls fun3

fun3()                                        d, e, f

fun1()                                        b, c, d

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun3().

   In fun3(), the local variables "d, e, f" of fun3() will be visible

   variable "b, c" of fun1() will be visible

   variable "a" of main() will be visible

c) main calls fun2; fun2 calls fun3; fun3 calls fun1

fun1()                                        b, c, d

fun3()                                        d, e, f

fun2()                                        c, d, e

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun1().

   In fun1(), the local variables "b, c, d" of fun1() will be visible

   variable "e, f" of fun3() will be visible

   variable "a" of main() will be visible

d) main calls fun1; fun1 calls fun3; fun3 calls fun2

fun2()                                        c, d, e

fun3()                                        d, e, f

fun1()                                        b, c, d,

main()                                        a, b,c

CALL STACK SHOWING THE VARIABLES OF EVERY FUNCTION

   From the above call stack diagram, it is very clear that the last function call is made to fun2().

   In fun2(), the local variables "c, d, e" of fun2() will be visible

   variable "f" of fun3() will be visible

     variable "b" of fun1() will be visible

   variable "a" of main() will be visible

The last function called will comprise of all its local variables and the variables other than its local variables from all its preceding function calls till the main function.

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Answer:

Option (D) is the correct answer of this question.

Explanation:

Moore's Law relates to Moore's theory that the number of transistors on a microchip doubles every two years while device costs are halved.

Interpretations  of Moore's law assert the computing power doubles every 18 months.The Moore's Law theory states that development is exponential.Moore's Law states that every couple of years we should expect our computers to increase their speed and capacity and we'll pay less for them.h

Option(A),Option(B) ,Option(c) and option(E) do not belongs to Moore's law  so these options are incorrect options.

5 0
1 year ago
Summary: Given integer values for red, green, and blue, subtract the gray from each value. Computers represent color by combinin
Dmitry_Shevchenko [17]

Answer:

Follows are the code to this question:

#include<iostream>//defining header file

using namespace std;// use package

int main()//main method

{

int red,green,blue,x;//declaring integer variable

cin>> red >>green>>blue;//use input method to input value

if(red<green && red<blue)//defining if block that check red value is greater then green and blue  

{

x = red;//use x variable to store red value

}

else if(green<blue)//defining else if block that check green value greater then blue  

{

x= green; //use x variable to store green value

}

else//defining else block

{

x=blue;//use x variable to store blue value

}

red -= x;//subtract input integer value from x  

green -=x; //subtract input integer value from x

blue -= x;//subtract input integer value from x

cout<<red<<" "<<green<<" "<<blue;//print value

return 0;

}

Output:

130 50 130

80 0 80

Explanation:

In the given code, inside the main method, four integers "red, green, blue, and x" are defined, in which "red, green, and blue" is used for input the value from the user end. In the next step, a conditional statement is used, in the if block, it checks red variable value is greater than then "green and blue" variable. If the condition is true, it will store red variable value in "x", otherwise, it will goto else if block.

  • In this block, it checks the green variable value greater than the blue variable value. if the condition is true it will store the green variable value in x variable.
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2 years ago
3.34 LAB: Mad Lib - loops in C++
Ganezh [65]

Answer:

A Program was written to carry out some set activities. below is the code program in C++ in the explanation section

Explanation:

Solution

CODE

#include <iostream>

using namespace std;

int main() {

string name; // variables

int number;

cin >> name >> number; // taking user input

while(number != 0)

{

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}

Note: Kindly find an attached copy of the compiled program output to this question.

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2 years ago
Suppose your name was George Gershwin. Write a complete program that would print your last name, followed by a comma, followed b
NNADVOKAT [17]

Answer:

I will write the code in C++ and JAVA      

Explanation:

<h2>JAVA CODE</h2>

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<h2>C++ Code:</h2>

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using namespace std;  

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6 0
2 years ago
Read 2 more answers
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Keith_Richards [23]

Answer:

Option a: a sorted array

Explanation:

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